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Elanso [62]
3 years ago
13

To test the effects of a new fertilizer, 100 plots were divided in half. Fertilizer A is randomly applied to one half, and B to

the other. This is ;
(a) an observational study.
(b) a matched pairs experiment.
(c) a completely randomized experiment.
(d) a block design, but not a matched pairs experiment.
(e) impossible to classify unless more details of the study are provided
Engineering
2 answers:
Kruka [31]3 years ago
7 0

Answer:

(b) a matched pairs experiment.

Explanation:

Matched pairs experiment are a special case of a random block design and is usually used when the experiment only has two conditions. In this case the conditions are the different fertilizers.

Oduvanchick [21]3 years ago
3 0

Answer: (b) a matched pairs experiment.

Explanation:

A match pair design or experiment is a randomized experimental procedure. It can be used in the experiment with two different sets of conditions and the subjects can be grouped into pairs. It is based on some of the blocking variable. The subjects are randomly assigned to different treatments.  

The given situation is an example of a matched pair experiment. This is because of the fact that there are two sets of conditions that is the use of fertilizer A and fertilizer B on different halves of the plot randomly to determine their effects.

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g The parameters of a certain transmission line operating at 휔휔=6 ×108 [rad/s] are 퐿퐿=0.35 [휇휇H/m], 퐶퐶=75 [pF/m], 퐺퐺=75 [휇휇S/m],
yKpoI14uk [10]

Explanation:

\begin{aligned}\gamma &=\sqrt{Z Y}=\sqrt{(R+j \omega L)(G+j \omega C)} \\&-\sqrt{|17|} j\left(6 \times 10^{8}\right)\left(0.35 \times 10^{-6}\right)|| 75 \times 10^{-6}\left|j\left(6 \times 10^{8}\right)\left(40 \times 10^{-12}\right)\right| \\&=0.094+j 2.25 \mathrm{m}^{-1}-\alpha+j \beta\end{aligned}

Therefore,

-\alpha-0.094 \mathrm{Np} / \mathrm{m} . \quad 3-2.25 \mathrm{rad} / \mathrm{m}, \text { and } \lambda-2 \pi / \beta-\underline{2.79} \mathrm{m}

Z_{0}-\sqrt{\frac{Z}{Y}}-\sqrt{\frac{R+j \omega L}{G+j \omega C}}-\sqrt{\frac{17+j 2.1 \times 10^{2}}{75 \times 10^{-6}+j 2.4 \times 10^{-2}}}-\frac{93.6-j 3.64 \Omega}{4}

5 0
3 years ago
A hollow pipe is submerged in a stream of water so that the length of the pipe is parallel to the velocity of the water. If the
Arlecino [84]

Answer:

increases by a factor of 6.

Explanation:

Let us assume that the initial cross sectional area of the pipe is A m² while the initial velocity of the water is V m/s², hence the flow rate of the water is:

Initial flow rate = area * velocity = A * V = AV m³/s

The water speed doubles (2V m/s) and the cross-sectional area of the pipe triples (3A m²), hence the volume flow rate becomes:

Final flow rate = 2V * 3A = 6AV m³/s = 6 * initial flow rate

Hence, the volume flow rate of the water passing through it increases by a factor of 6.

8 0
3 years ago
5. A typical paper clip weighs 0.59 g and consists of BCC iron. Calculate (a) the number of
marta [7]

Answer:

(a) 3.185*10^{21} cells

(b) 6.37*10^{21} atoms

Explanation:

(a)

Volume, V of unit cell

V=(2.866*10^{-8})^{3}=2.354*10^{-23}

Number of unit cells, N

N=\frac {W_{mat}}{V\rho_{mat}} Where W_{mat} is weight of material and \rho_{mat} is density of material

N=\frac{0.59}{7.87*(2.354*10^{-23}}=3.185*10^{21} cells

(b)

Number of atoms in paper clip

This is a product of number of unit cells and number of atoms per cell

Since iron has 2 atoms per cell

Number of atoms of iron=3.185*10^{21} cells*2 atoms/cell=6.37*10^{21} atoms

8 0
3 years ago
A carbon resistor has a resistance of 976 ohms at 0 degrees C. Determine its resistance at 89 degrees C​
nignag [31]

Answer:

1028.1184 Ohms

Explanation:

<u>Given the following data;</u>

  • Initial resistance, Ro = 976 Ohms
  • Initial temperature, T1 = 0°C
  • Final temperature, T2 = 89°C

Assuming the temperature coefficient of resistance for carbon at 0°C is equal to 0.0006 per degree Celsius.

To find determine its new resistance, we would use the mathematical expression for linear resistivity;

R_{89} = R_{0} + R_{0}(\alpha T)

Substituting into the equation, we have;

R_{89} = 976 + 976*(0.0006*89)

R_{89} = 976 + 976*(0.0534)

R_{89} = 976 + 52.1184

R_{89} = 1028.1184 \ Ohms

5 0
3 years ago
Hello , how are yall:))))
SVEN [57.7K]

Answer:

eh I'm good hbu?????????

6 0
3 years ago
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